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Tag 0629

Chapter 15: More on Algebra > Section 15.26: The Koszul complex

Lemma 15.26.8. Let $R$ be a ring. Let $f_1, \ldots, f_r$ be a sequence of elements of $R$. The complex $K_\bullet(f_1, \ldots, f_r)$ is isomorphic to the cone of the map of complexes $$ f_r : K_\bullet(f_1, \ldots, f_{r - 1}) \longrightarrow K_\bullet(f_1, \ldots, f_{r - 1}). $$

Proof. Special case of Lemma 15.26.7. $\square$

    The code snippet corresponding to this tag is a part of the file more-algebra.tex and is located in lines 5653–5664 (see updates for more information).

    \begin{lemma}
    \label{lemma-cone-koszul}
    Let $R$ be a ring. Let $f_1, \ldots, f_r$ be a sequence of elements
    of $R$. The complex $K_\bullet(f_1, \ldots, f_r)$ is isomorphic to the
    cone of the map of complexes
    $$
    f_r :
    K_\bullet(f_1, \ldots, f_{r - 1})
    \longrightarrow
    K_\bullet(f_1, \ldots, f_{r - 1}).
    $$
    \end{lemma}
    
    \begin{proof}
    Special case of
    Lemma \ref{lemma-cone-koszul-abstract}.
    \end{proof}

    Comments (2)

    Comment #2447 by Raymond Cheng on March 9, 2017 a 2:42 am UTC

    Should be $f_r \colon K_\bullet(f_1,\ldots,f_{r-1}) \to K_\bullet(f_1,\ldots,f_{r-1})$ instead of $f_n\colon K_\bullet(f_1,\ldots,f_{r-1}) \to K_\bullet(f_1,\ldots,f_{r-1})$.

    Comment #2489 by Johan (site) on April 13, 2017 a 10:57 pm UTC

    Thanks, fixed here.

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